Is frequency inversely proportional to wavelength Squared? In our QM intro class our professor was introducing us to concepts of energy, momentum, frequency, wavelength etc of photons an electrons, On one side of the board he had the various expressions for these things and the properties etc. Then on the other side of the board he wrote this equation $\omega=\frac{\hbar\pi}{m}.(\frac{1}{\lambda})^2$ which was confusing, because the left side is frequency so this means $\omega=\text{constant}(\frac{1}{\lambda})^2$ so why is frequency is inversely proportional to wavelength squared, Why it's confusing is because we know of the relation speed$=f\lambda$ so frequency and wavelength are inversely related but this equation says frequency is inversely related to the squared wavelength, 1. I would like to know where this equation comes from and 2. does it mean frequency of photons are sometimes inversely proportional to wavelength squared or just wavelength (is equation wrong with a typo - should be $\frac{1}{\lambda}$? Thanks for your help. PS A student put their hand up and asked about this equation telling the professor it's wrong and should be $\frac{1}{\lambda}$ and the professor said "NO! It's right and you can derive it"- but we all tried and couldnt. I've also checked this site and I online seacrhed it too but couldn't find anything about it. Many thanks for any help.
 A: The relation $c=f\lambda$ does indeed hold for light. The equation you have (the derivation is below) holds for matter waves that are nonrelativistic - or moving with low velocity.
For more, see De Broglie's hypothesis
where it is explained how matter has wave-like characteristics like an electron which has mass.
I think your professor is trying to point out that for matter waves, frequency actually is inversely proportional to the square of the wavelength, whereas for light the relation $f= \frac{c}{\lambda}$ or $\omega=\frac{2\pi c}{\lambda}$ holds.
For a single nonrelativistic particle with mass, like a low velocity electron,
its energy and momentum
can be described by $$E=\frac 12 mv^2\ \ \text{and}
\ \ p=mv\rightarrow E=\frac{p^2}{2m}$$ This means we can write $$h\nu=\frac{p^2}{2m} \
\rightarrow \nu=\frac{p^2}{2mh}$$ From the link above, the de Droglie wavelength $\lambda=\frac hp$ or $p=\frac{h}{\lambda}$ so  $$\nu=\frac{h}{2m}\cdot\frac{1}{\lambda^2}$$ and since $\omega=2\pi\nu$ we can write
$$\boxed{\omega=\frac{\pi h}{m}\left(\frac{1}{\lambda^2}\right)}$$
I can see in your equation there is the reduced Planck's constant $\hbar$ but here it is
just Planck's constant. That must be a typo on your's or your professor's part.
Remember the important lesson here of which your professor is drawing your attention to:
For light, frequency and wavelength
are inversely proportional, while for matter waves frequency is inversely proportional to the square of the wavelength.
A: Your professor is right, concerning the wavelength and frequency of a nonrelativistic matter wave. Frequency is proportional to energy, which is proportional to momentum squared. Meanwhile, momentum is inversely proportional to wavelength. More explicitly:
$$E=\hbar\omega=\frac{p^2}{2m}=\frac{(\hbar k)^2}{2m}$$
$$\implies\omega=\frac{\hbar}{2m}k^2$$
To put this in terms of wavelength instead of wavenumber, we use $\lambda=\frac{2\pi}{k}$:
$$\implies\omega=\frac{2\hbar\pi^2}{m\lambda^2}$$
For photons, we may no longer assume that $E=\frac{p^2}{2m}$. For photons, the relevant relation between $\omega$ and $\lambda$ is instead:
$$\omega=\frac{2\pi c}{\lambda}$$
